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What would the minimum work function for a metal have to be for visible light \((380-750 \mathrm{nm})\) to eject photoelectrons?

Short Answer

Expert verified
The minimum work function is approximately 1.65 eV.

Step by step solution

01

Understand the Problem

The problem is asking for the minimum work function of a metal that allows photoelectric effect from visible light, which ranges from 380 nm to 750 nm. The work function is the minimum energy needed to eject electrons from the metal when light shines on it.
02

Use the Energy-Wavelength Relationship

To find the energy corresponding to the minimum and maximum wavelengths of visible light, use the formula: \[ E = \frac{hc}{\lambda} \]where \( h \) is Planck's constant \( (6.626 \times 10^{-34} \text{J s}) \), \( c \) is the speed of light \( (3 \times 10^8 \text{ m/s}) \), and \( \lambda \) is the wavelength. We'll use this for \( \lambda = 750 \text{ nm} \) (longest wavelength, so minimum energy).
03

Convert Wavelength to Meters

Convert 750 nm to meters: \[ 750 \text{ nm} = 750 \times 10^{-9} \text{ m} \]
04

Calculate Minimum Energy

Substitute \( h \), \( c \), and \( \lambda \) into the energy formula: \[ E = \frac{6.626 \times 10^{-34} \text{ J s} \times 3 \times 10^8 \text{ m/s}}{750 \times 10^{-9} \text{ m}} \]\[ E \approx 2.65 \times 10^{-19} \text{ J} \]
05

Convert Energy to Electronvolts

Convert the energy from joules to electronvolts, using the conversion factor \( 1 \text{ eV} = 1.602 \times 10^{-19} \text{ J} \):\[ 2.65 \times 10^{-19} \text{ J} \times \frac{1 \text{ eV}}{1.602 \times 10^{-19} \text{ J}} \approx 1.65 \text{ eV} \]
06

Conclusion

The minimum work function for the metal to allow the photoelectric effect from visible light is approximately 1.65 eV. This corresponds to the longest wavelength of 750 nm.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

work function
In the context of the photoelectric effect, the work function is a crucial concept. It refers to the minimum amount of energy required to eject an electron from the surface of a metal. This energy barrier must be overcome by the energy of the incoming photons for photoelectrons to be emitted.

When light shines on a metal, only photons with energy equal to or greater than the work function can release electrons. This principle is fundamental in understanding various technologies, such as solar panels, where electrons are liberated to generate electricity.

The work function is specific to each material and usually expressed in electronvolts (eV). Computation of the work function often involves converting photon energy from joules to electronvolts. This conversion provides a tangible scale for measuring and comparing the work functions of different materials.
visible light
Visible light is a spectrum of electromagnetic radiation that the human eye can perceive. It ranges from about 380 nm (violet) to 750 nm (red) in wavelength. This spectrum represents a small portion of the electromagnetic spectrum, but it's critical for numerous applications and phenomena.

In the photoelectric effect, visible light can play a key role in ejecting electrons from materials. However, not all visible light has enough energy to cause this effect. Only light within certain wavelengths—and therefore energies—can potentially have photons that meet or exceed the metal's work function.
  • Shorter wavelengths, like violet light, have higher energy photons.
  • Longer wavelengths, like red light, have lower energy photons.
Understanding the relationship between wavelength and photon energy is vital for harnessing visible light in technologies such as cameras and solar cells.
wavelength-energy relationship
The wavelength-energy relationship is fundamental to the concept of the photoelectric effect. This relationship is described by the formula:
\[ E = \frac{hc}{\lambda} \]where:
  • \( E \) = energy of the photon,
  • \( h \) = Planck's constant \( (6.626 \times 10^{-34} \text{Js}) \),
  • \( c \) = speed of light \( (3 \times 10^8 \text{ m/s}) \),
  • \( \lambda \) = wavelength of the light.
This equation explains why shorter wavelengths correspond to higher energy photons, which are more capable of overcoming the work function of metals.

For instance, in the original problem context, the longest wavelength of visible light (750 nm) was considered to calculate the minimum energy. Such applications are essential in fields like quantum mechanics and optoelectronics, where understanding the precise interaction of light and matter can lead to technological advancements.

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Most popular questions from this chapter

When ultraviolet light with a wavelength of 400.0 \(\mathrm{nm}\) falls on a certain metal surface, the maximum kinetic energy of the emitted photoelectrons is measured to be 1.10 eV. What is the maximum kinetic energy of the photoelectrons when light of wavelength 300.0 nm falls on the same surface?

BIO A laser used to weld detached retinas emits light with a wavelength of 652 \(\mathrm{nm}\) in pulses that are 20.0 \(\mathrm{ms}\) in duration. The average power during each pulse is 0.600 \(\mathrm{W}\) (a) How much energy is in each pulse in joules? In electron volts? (b) What is the energy of one photon in joules' In electron volts? (b) How many photons are in each pulse?

An x-ray photon is scattered from a free electron (mass \(m )\) at rest. The wavelength of the scattered photon is \(\lambda^{\prime},\) and the final speed of the struck electron is \(v\) . (a) What was the initial wave-length \(\lambda\) of the photon? Express your answer in terms of \(\lambda^{\prime}, v\) and \(m .\) (Hint. Use the relativistic expression for the electron kinetic energy.) (b) Through what angle \(\phi\) is the photon scattered? Express your answer in terms of \(\lambda, \lambda^{\prime},\) and \(m\) . (c) Evaluate your results in parts (a) and (b) for a wavelength of \(5.10 \times 10^{-3}\) nm for the scattered photon and a final electron speed of \(1.80 \times 10^{8} \mathrm{m} / \mathrm{s}\) . Give \(\phi\) in degrees.

A photon scatters in the backward direction \(\left(\phi=180^{\circ}\right)\) from a free proton that is initially at rest. What must the wavelength of the incident photon be if it is to undergo a 10.0\(\%\) change in wavelength as a result of the scattering?

(a) If the average frequency emitted by a 200 -W light bulb is \(5.00 \times 10^{14} \mathrm{Hz}\) , and 10.0\(\%\) of the input power is emitted as visible light, approximately how many visible-light photons are emitted per second? (b) At what distance would this correspond to \(1.00 \times 10^{11}\) visible-light photons per square centimeter per second if the light is emitted uniformly in all directions?

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